Area bounded by the curve y = min {sin^2x, co | vedclass

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(C) The curve $y = \min \{\sin^2x, \cos^2x \}$ changes its definition at points where $\sin^2x = \cos^2x$,i.e.,$	an^2x = 1$,which gives $x = \frac{\p

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$\frac{5(\pi - 2)}{8}$ square units

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(C) The curve $y = \min \{\sin^2x, \cos^2x \}$ changes its definition at points where $\sin^2x = \cos^2x$,i.e.,$	an^2x = 1$,which gives $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}$.In the interval $[0, \frac{\pi}{4}]$,$\sin^2x \leq \cos^2x$,so $y = \sin^2x$.In the interval $[\frac{\pi}{4}, \frac{3\pi}{4}]$,$\cos^2x \leq \sin^2x$,so $y = \cos^2x$.In the interval $[\frac{3\pi}{4}, \frac{5\pi}{4}]$,$\sin^2x \leq \cos^2x$,so $y = \sin^2x$.The total area $A$ is given by:$A = \int_{0}^{\frac{\pi}{4}} \sin^2x \, dx + \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cos^2x \, dx + \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} \sin^2x \, dx$.Using $\sin^2x = \frac{1-\cos 2x}{2}$ and $\cos^2x = \frac{1+\cos 2x}{2}$:$A = \int_{0}^{\frac{\pi}{4}} \frac{1-\cos 2x}{2} dx + \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1+\cos 2x}{2} dx + \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{1-\cos 2x}{2} dx$.$A = \left[ \frac{x}{2} - \frac{\sin 2x}{4} ight]_{0}^{\frac{\pi}{4}} + \left[ \frac{x}{2} + \frac{\sin 2x}{4} ight]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} + \left[ \frac{x}{2} - \frac{\sin 2x}{4} ight]_{\frac{3\pi}{4}}^{\frac{5\pi}{4}}$.$A = (\frac{\pi}{8} - \frac{1}{4}) + ((\frac{3\pi}{8} - \frac{1}{4}) - (\frac{\pi}{8} + \frac{1}{4})) + ((\frac{5\pi}{8} - 0) - (\frac{3\pi}{8} + \frac{1}{4}))$.$A = \frac{\pi}{8} - \frac{1}{4} + \frac{2\pi}{8} - \frac{2}{4} + \frac{2\pi}{8} - \frac{1}{4} = \frac{5\pi}{8} - 1 = \frac{5\pi - 8}{8}$.

Area bounded by the curve $y = \min \{\sin^2x, \cos^2x \}$ and $x-$ axis between the ordinates $x = 0$ and $x = \frac{5\pi}{4}$ is

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